Question

Fit a cubic trend function to the data (-2,7), (-1,9), (0,9), (1,8), (2,7). Let $p_0(t) = 1$, $p_1(t) = t$ and $p_2(t) = t^2 - 2$. The orthogonal cubic polynomial is $p_3(t) = \frac{5}{6}t^3 - \frac{17}{6}t$. The vectors for $p_0$, $p_1$, and $p_2$ are $\begin{bmatrix} 1\\1\\1\\1\\1 \end{bmatrix}$, $\begin{bmatrix} -2\\-1\\0\\1\\2 \end{bmatrix}$ and $\begin{bmatrix} 2\\-1\\-2\\-1\\2 \end{bmatrix}$, respectively. $\hat{p}(t) = $

          Fit a cubic trend function to the data (-2,7), (-1,9), (0,9), (1,8), (2,7). Let $p_0(t) = 1$, $p_1(t) = t$ and $p_2(t) = t^2 - 2$. The orthogonal cubic polynomial is $p_3(t) = \frac{5}{6}t^3 - \frac{17}{6}t$. The vectors for $p_0$, $p_1$, and $p_2$ are $\begin{bmatrix} 1\\1\\1\\1\\1 \end{bmatrix}$, $\begin{bmatrix} -2\\-1\\0\\1\\2 \end{bmatrix}$ and $\begin{bmatrix} 2\\-1\\-2\\-1\\2 \end{bmatrix}$, respectively. 
$\hat{p}(t) = $

Fit a cubic trend function to the data (-2,7), (-1,9), (0,9), (1,8), (2,7). Let p0(t) = 1, p1(t) = t and p2(t) = t^2 - 2. The orthogonal cubic polynomial is p3(t) = (5)/(6)t^3 - (17)/(6)t. The vectors for p0, p1, and p2 are < b m a t r i x >, < b m a t r i x > and < b m a t r i x >, respectively.
p̂(t) =

Added by Joseph J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Angelo Rendina

08/21/2019

Step 1

Step 1: The orthogonal cubic polynomial is p₃(t) = -5/6 t³ - 17/6 t. Show more…

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Fit a cubic trend function to the data (-2,7), (-1,9), (0,9), (1,8), (2,7). Let p₀(t) = 1, p₁(t) = t and P₂(t) = t² - 2. The orthogonal cubic polynomial is p₃(t) = -5/6 t³ - 17/6 t. The vectors for p₀, p₁, and p₂ are [1] [1] [1] [1] [1] [-2] [-1] [0] [1] [2] [2] [-1] [-2] [-1] [2] , respectively. p(t) =